Q29E Calculate the moment of inertia ... [FREE SOLUTION]

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University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition 路 1596 pages

ISBN: 9780321973610

Chapter 1: Q29E (page 296)

Calculate the moment of inertia of each of the following uniform objects about the axes indicated. Consult Table 9.2 as needed.
(a) A thin 2.50-kg rod of length 75.0 cm, about an axis perpendicular to it and passing through (i) one end and (ii) its center, and (iii) about an axis parallel to the rod and passing through it.
(b) A 3.00-kg sphere 38.0 cm in diameter, about an axis through its center, if the sphere is (i) solid and (ii) a thin-walled hollow shell.
(c) An 8.00-kg cylinder, of length 19.5 cm and diameter 12.0 cm, about the central axis of the cylinder, if the cylinder is (i) thin-walled and hollow, and (ii) solid.

Short Answer

Expert verified

(a) For rod, we will use equations from table 9.2.
(b) Thus, the solid is .

(ii) Thus, the thin-walled is $0.072 k g . m^{2}$ .

Step by step solution

Step:-1 explanation

Mass of the rod = 2.5 kg

Length of the rod = 75 cm

$= \frac{75}{100} = 0.75 c m l = \frac{1}{3} M L^{2}$

Step:-2 solution

Put the value here,

$= \frac{1}{3} (2.5 脳 0.75) = 0.469 k m . m^{2}$

Mass of the rod = 2.5 kg

Length of the rod = 75 cm

$= \frac{75}{100} = 0.75 m l = \frac{1}{12} M L^{2}$

Putting the value here,

$= \frac{1}{12} (2.5 脳 0.75) = 0.117 k m . m^{2}$

For a very thin rod, since the masses are near to the axis, thus the moment of inertia =0

Step:-3 solid

given M =3.00 kg

$d = 38.0 c m r = \frac{d}{2} = \frac{38}{2} = 19 c m$

given M =3.00kg

d =38.0 cm

Step:-4 Concept

$\begin{aligned} I = \frac{1}{2} 脳 3 脳 (19)^{2} \\ = \frac{2}{5} 脳 3 脳 361 \\ = 0.4 脳 3 脳 361 \\ = 433.2 \\ I = 0.043 kg 鈰� m^{2} \end{aligned}$

Step:-5 thin-walled hollow shell

given M =3.00 kg

$d = 38.0 c m r = \frac{d}{2} = \frac{38}{2} = 19 c m$

Step:-6 calculation

$l = \frac{2}{3} M R^{2} = \frac{2}{3} M {(\frac{D}{2})}^{2} = 0.072 k g . m^{2}$

Step:-7 explanation

$d = 12 r = \frac{d}{2} r = 6$

Step:-8 Solution

$l = \frac{1}{2} M R^{2}$

Put the value here

$l = M {(\frac{D}{2})}^{2} = 0.0288 k g . m^{2}$

Thus, the answer $0.0288 k g . m^{2}$ .

Step 9:-explanation

$d = 12 r = \frac{d}{2} r = 6$

Step:-10 solution

$l = \frac{1}{2} M {(\frac{D}{2})}^{2} = \frac{1}{2} 脳 8 脳 6^{2} = \frac{8 脳 36}{2} = 144 = 0.0144 k g . m^{2}$

Hence, the solution is $0.0144 k g . m^{2}$ .

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Short Answer

Step by step solution

Step:-1 explanation

Step:-2 solution

Step:-3 solid

Step:-4 Concept

Step:-5 thin-walled hollow shell

Step:-6 calculation

Step:-7 explanation

Step:-8 Solution

Step 9:-explanation

Step:-10 solution

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